I want to generalize the result for $\mathbb{R}^n$, that any non-trivial open set has positive measure.
Open sets and measure zero
Question: For which types of topological measure spaces is it true that every non-trivial (i.e. non-empty) open set has positive measure?
My thinking was that only inner regularity of the measure, combined with local compactness of the space, would be necessary.
There are some problems with this:
(1) We don't just need that every point has a compact neighborhood for this to work -- otherwise we would be able to prove that any set containing a compact set has positive measure.
We need that every point has a compact neighborhood with positive measure. This does not seem to follow from the given candidate assumptions. And
(2) The possible definitions of local compactness vary depending on whether or not the space is Hausdorff. Obviously the topological measure space needs to be Hausdorff in order for it to be inner regular, but there are weaker notions which might be sufficient and which don't require $T_2$.
Apparently we also need left-invariance, i.e. that our topological measure space is in fact a measurable topological group. measure of open set with measure Haar But from the answer given to that question the following is still unclear to me:
Do we really need Hausdorff? And do we really need group structure and left-invariance to conclude that every point has a compact neighborhood with positive measure?
One question: doesn't a Borel measure just have to be defined on open sets? Why does it need to have positive measure for all open sets?
– Chill2Macht Aug 05 '16 at 19:07